For the uninitiated, a jellybean guessing contest (or jellybean counting contest) is a game held at fairs or parties or anyplace, where someone fills a jar or other large, typically transparent container with a known quantity of jellybeans and gives a prize to whomever guesses closest to the actual number. Participant estimates often vary by several orders of magnitude but I believe you can increase your odds and minimize the luck involved with a little knowledge (they say a little knowledge is a dangerous thing, but hey, it’s just a jelly bean contest).

Strategies for Victory

There are a couple of strategies you may use in calculating the number of jellybeans in a jar or other container. In order of increasing complexity, they are:

1) Guess

It’s just a game. Arm yourself with the knowledge that there are 930 jelly beans in a US gallon (about 245 / litre) and venture a WAG. I suspect you wouldn’t be reading this if that’s all you were after, loser. So…

2) Count them

Don’t count each bean, of course, but if you are allowed to lift the container, you can count the number of jelly beans in one row (remember to compensate for tapering at the bottom of the container) and multiply that number times the number of jellybeans the container is tall. This will yield a very good estimate.

3) Equate them

This strategy is best if you are “guessing” the number of jellybeans in a known volume. In fact, it’s almost fool-proof in that situation. You need to know:

a) The approximate volume of one jelly bean can be thought of as a small cylinder 2 cm long and 1.5 cm in diameter (Precisely articulated as: Volume of 1 Jelly Bean = h(pi)(d/2)^2 = 2cm x 3 (1.5cm/2)^2 = 3.375 or 27/8 cubic centimeters)

b) Due to the Jelly Bean spape and irregularities, there is considerable airspace in the container, along with the jelly beans. It can be assumed that 20% of a given volume is air rather than jellybeans (though for very small or irregularly shaped containers, this figure might be slightly more… never estimate more than 25% air by volume. Really 20% is the best value to use for n > 100)

So, to get your answer, you will want to determine the number of cubic centimeters in the container volume and multiply that number by can simply use a calculator to divide the volume of the container in cubic centimeters by 2.7 (3.375 * .8 to allow for air space). Google is a great too for doing all of this. For example a search for “cubic centimeters per gallon” and Google returns “1 US gallon = 3 785.4118 cubic centimeters”. You can then use Google (they call it Google Calculator but you use the normal search engine box) to calculate your answer. 5 gallon bucket = 5 x 3785.4 = 18,927. You just pop ’round the gorn-and-scumbles, and, Jack’s a doughnut, there you are! edit: MK in the comments gave the following tip which should make this easier and more accurate:

Instead of trying to calculate the volume of the jar, one could try what I did – Look for the brand of the jar (if it is a branded one) and google for its volume instead. If the jar is of the big-brand kind, you can usually find its dimensions & capacity info either on their e-stores, or mentioned in how-to guides (think crafting, food-storing blogs) online.

What if I’m Stuck?

But what to do in the inevitable scenario of a container that you aren’t allowed to touch and of an unknown volume? For the cowards and weaklings, there’s always option one above. For the stout-hearted, you will need to improvise. The best way is to try to use option 3 with an estimated volume. Does the container look about the size of any containers of a known volume? A Gallon of milk? A two litre bottle? A 5 gallon bucket? Containers tend to be an exact volume… 750 ml, 2 litres, etc. If it looks pretty close, it probably is.

If it isn’t close enough, or if you just want to double check your conclusion, you can attempt to calculate the volume by sight. As we already mentioned above, for a cylindrical object, volume = (area of the base) * height = π r^2 h. You will need to estimate both r and h of that equation. Note that r (radius) is HALF the diameter. A useful device for helping with length estimates is a sheet of paper. Various papers like signup sheets or flyers or whatever are often sitting near the container in question. You can use the known size of the paper (8 1/2″ x 11″) as a comparison for the container.

Calculating the volume of a non-cylindrical contianer is a much more challenging feat. If the shape is near recatangular, you can obviously just do length x height x width. If it is nearly spherical, you can use V = 4πr3 / 3. As the shape gets more interesting, your best best is to attempt to guess at the volume and then estimate. You’ll still be at an advantage with this preparation, but it isn’t nearly as great as if you knew the volume. If you are serious about estimating the volume of more odd-shaped containers, see spheroid and ellipsoid at Wikipedia.

The End?

The final dilemma is how to handle the glory, honor and fame that inevitably accompanies jelly bean contest victory. Should you gloat or fake humble? Mock the losers or encouragingly pat shoulders? These questions are really beyond the scope of this treatise, but I wanted to send you off forewarned and prepared for the crazed paparazzi to come. Good luck and good night.

Jetasaid,

Hm! If you want to make up for air around your ideal bean-shaped container, you want to assume a larger jellybean, not a smaller one. Without compensation, you calculated 3.375; assuming 20% dead air, 80% jelly beans, and we’re looking at 3.375 : X = 80% : 100%, or X = 100 * 3.375 / 80, or 4.21 ccm per bean.

Wow, that’s a large jelly bean. But it is slightly closer to the number I get if I divide one gallon – about 3785 ccm – by the 930 jellybean/gallon figure you mentioned: 4.0698.

Really, it would be much easier if google just added jellybeans as a unit into their calculating module!

Joel Beansaid,

ditto. This guy is right. At first I just passed up this comment thinking that the calculation came out the same whether you multiply by .8 before or after dividing the volume of the container by the volume of a jelly bean. But in order for that to be true, you would need to multiply the numerator (volume of the container) by .8, not the denominator (volume of the jelly bean). In your calculation, you’re multiplying the volume of the jelly bean by .8, which is wrong.

Thanks for the article though! It makes a lot more sense than those ones out there saying that theres 5,000 jelly beans in a gallon! I’ve got a candy counting contest later today and hopefully this will help. But I’m not sure if it will be jelly beans.

Kevinsaid,

Élisabethsaid,

Hello there. Well, first, I have to say, most of the comments are really nerdy. and second, I have totally been obsessing about these contests and am probably a bigger nerd that all of you. I have to menton that in the enventuallity that the jar is a shere with flat sides of 1gallon there are only 665 about that fit. you have to think about how the angle of the side will enfluence the air/jelly bean ratio. just a thought.

Ben Doversaid,

David Muddsaid,

My tech physics classes have to solve a Fermi Question concerning the number of jelly beans in a small box- using two methods. 1st – estimating the number much the way you’ve done with the simple geometry of a cylinder…the problem then is the air space. Our estimates ranged from zero (groups that don’t think at all) to a high of 50% (groups that don’t think well). I require them to discover a simple way to test their air space estimate – the 2 favorite ways are to pour sand into the box after the jelly beans have been added then sift the mixture and to lay out a layer of jelly beans and lightly spray paint over them and look at the paint shadow. The 2nd estimate technique is to build a smaller model of the box, say 2 cubic centimeters, and fill it with jelly beans. Neither are tremendously accurate but both ways allow students to discover methods estimation.

Nice Catchsaid,

no offense, but if the dude plagiarized from a published source; that really is a very frowned upon offense.

I mean no one died in the stealing of Dave Barry’s words, but for real, he shouldn’t pawn them off as his own, and God forbid someone wanted to annotate a paper, using THIS article as the source, then THEY could get kicked out of school for stealing Dave Barry’s words, when in fact, this article was the thief.

while maybe he should get outside, he was absolutely right to point out the stealing of someone else’s thoughts and words. Ropes, Shame on you that you can’t see the offense. Perhaps you’re one who buys their book reports online?!?

get a life???said,

ok first of all WOW its a jellybean contest…. just guess if uwin u win like 5 dollars dang your 5 dollars richer. U know what i thnk u could do with that 5 dollars go buy your self a fkin 5 dolar footlong.

Mollykinssaid,

Wow people. I quoted him. Didn’t steal his comment to claim as my own. It’s a funny line. I quote movies, books, songs all the time. Quoting one line, particularly a nonsensical line for comedic relief isn’t wrong or even offensive IMHO.

Quotation marks around the quoted material indicate clearly the material from another. In addition, attribution is also an ethical requirement. For example, you might have said, “As Dave Barry once wrote, ‘You just pop ’round the gom-and-scumbles, and, Jack’s a doughnut, there you are!'”

Percysaid,

cheekysaid,

There is a way to get a more accurate guesstimate, buy lots of jelly beans fill various containers , weigh several beans and work out the average weight of an individual bean, make a note of how many in various containers, thus if presented with a larger or slightly smaller container than you have noted, then you will know the max/min of what is contained, also to save some cash use a dried bean alternative to work it out

[…] right answer; I’ve seen it happen a few times and it’s a genius technique. Additional perspectives are available but I like the groupthink route the best. This entry was posted in School, Thought. […]

Deansaid,

Deansaid,

[…] for a big number because they are risky and overconfident? Reasonably confident in some rational calculation regarding the spatial relationship between the jar and the beans? All of these kinds of guesses are likely, and to some degree governed by an emotional make up of […]

georgesaid,

In regards to the voids, which in not unlike soils engineering (of which has several equations to calculate the voids), a simple test would be to fill a large jar (of known volume) with jelly beans to the rim, place a lid on the jar with holes in it, submerse the jar into a tub of water, allow the jar to fill with water, and then pull out the jar and quickly empty the jar of water into another dry container of which you could then measure the volume of water which should provide an estimate of the void. Of course, this would all need to be done rather quickly as the jelly bean will begin to dissolve once water is introduced.

Lorensaid,

“determine the number of cubic centimeters in the container volume and multiply that number by can simply use a calculator to divide the volume of the container in cubic centimeters by 2.7” has a glaring omission. What is getting multiplied times the volume before the division by 2.7?

Charlie Brownsaid,

Aransaid,

Someone needs to buy a few packets of jellybeans and do some testing for us! Once the jellybeans are added in, fill the rest of the container with water to the rim, then drain the water and measure how much water was used to fill the air, then that can be used as a % of wasted jellybean space.

Charlie Brownsaid,

Charlie Brownsaid,

thank you soooo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!! because of you i won an xbox 360 along with 1600 ice cream flavored jellybeans. i am NOT dumb but i’ve never been good at jellybean contests but now i can guess the EXACT amount 🙂

P.S my name isnt charlie brown i just didnt want my name flashing all over the internet

Charlie Brownsaid,

by the way if you want to know what pi = then here are the first 124 digits
(out of ∞ of course!) π=3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938…

Charlie Brownsaid,

ardisaid,

[…] How to Win a Jellybean Guessing Contest « Cleverness: Getting …Mar 7, 2007 … For the uninitiated, a jellybean guessing contest (or jellybean counting contest) is a game held at fairs or parties or anyplace, where someone … […]

[…] decent ones. Not too irrational, right? I enjoy winning. Whether it be sports, arguments, gambling, guessing how many jelly beans are in the mason jar, or finding Waldo. I especially enjoy saving money. Perhaps, a close second behind finding […]

Mastersaid,

Another way: Wait until some other people have guessed. Add their guesses and divide the sum by the amount of guesses. The more people have already guessed, the better. That’s the intellegence of the masses ;o)

opsaid,

[…] one of those jellybean counting contests. But I probably could come close to winning if I took some time to follow the formula. I will keep that in mind for next time. Oof? Next time? What am I saying? I don’t think […]

Mollykinssaid,

MKsaid,

So thought i would share a tip that may help. Instead of trying to calculate the volume of the jar, one could try what I did – Look for the brand of the jar (if it is a branded one) and google for its volume instead. If the jar is of the big-brand kind, you can usually find its dimensions & capacity info either on their e-stores, or mentioned in how-to guides (think crafting, food-storing blogs) online.

That said, I did do some physical measuring of the office’s jar to verify against the internet’s info to make sure i got the correct jar. 🙂

bradsaid,

Shawntae Sharpesaid,

Emily Bsaid,

For cylindrical containers, I count the # of jelly beans around the circumference and for the height. Then I use the 22/7 version of pi (not exact but it’s close enough for our purposes) to find the diameter (unless you’d like to count the diameter; I couldn’t) and then I find the volume of a cylinder with the equation V = πr2h

For squarish ones, I do the same thing. Count the # of beans on each unique side, multiply for the 2D area and then multiply that by the # of jelly beans tall. Voila! It’s usually enough to get you close enough to win, depending on the rules.

I’m impressed, I must say. Really rarely do I encounter a blog that’s both educative and entertaining, and let me tell you, you have hit the nail on the head. Your idea is outstanding; the issue is something that not enough people are speaking intelligently about. I am very happy that I stumbled across this in my search for something relating to this.

JEWELL BROWN-TREE-JB....said,

I’m surprised no-one has mentioned averaging out the guesses if you have access to the recorded guesses – the more the better. Having time to sit down to do the averaging is unlikely but sometimes can be done. What brought me here to this article was attempting to understand how the phenomenon of averaging works in this situation to be so accurate. Not that it is directly relevant to the purpose of the article, but I believe it is related to quantum mechanics and the slit experiments.